如何按特定间隔遍历字符串？求修正统计'bob'数量的Python代码

Got it, let's break down what's going wrong with your current code and get it working properly.

What's Off with the Original Code? #

• You're skipping potential matches: Using range(0, s_len+1, 3) means you only check every 3rd character as the start of "bob". But "bob" can start at any index—like in "bobob", there's a "bob" starting at index 0 and index 2, but your loop would never check index 2.
• You're checking the whole string every time: The line if bob_counter in s doesn't look at the current position in the loop—it just checks if "bob" exists anywhere in the string. So for your example "waterbob", the loop runs 3 times (indices 0, 3, 6) and adds 1 each time, even though there's only one "bob" total.
• The else: break is unnecessary: This stops the loop as soon as it hits a position that doesn't have "bob", which would miss any matches later in the string.

Corrected Code #

Here's a fixed version that counts every single "bob" correctly, even overlapping ones:

def bobs_counter():
    bob_count = 0
    s = 'waterbob'
    bob_target = "bob"
    target_length = len(bob_target)
    string_length = len(s)
    
    # Loop through every valid starting index for "bob"
    for i in range(string_length - target_length + 1):
        # Check if the 3-character chunk starting at i is exactly "bob"
        if s[i:i+target_length] == bob_target:
            bob_count += 1
    
    print(bob_count)

bobs_counter()

What Changed (and Why)? #

• Valid starting indices only: range(string_length - target_length + 1) makes sure we only check positions where there are enough characters left to form "bob". For your 8-character string, that means indices 0 to 5 (since 5 + 3 = 8—no need to check indices 6 or 7, because you can't fit 3 characters there).
• Check the exact substring: Instead of scanning the whole string each time, we grab the 3 characters starting at the current index (s[i:i+target_length]) and compare it directly to "bob". This way, we only count matches at that specific position.
• Dynamic length: Using len(bob_target) instead of hardcoding 3 makes the function flexible—if you ever want to count a different substring later, you just change bob_target.

Test It Out #

For your example "waterbob", this code will correctly print 1. If you test it with "bobbob", it'll return 2, and with "bobob" it returns 2 (catching both the "bob" at index 0 and index 2).

内容的提问来源于stack exchange，提问作者user9473137